Problem 46 - College Algebra

Question

Solve the system for \(x, y,\) and \(z\). $$ \begin{aligned} x+y+z &=3 \\ \frac{x-1}{2}+\frac{y-3}{2}+\frac{z+1}{2} &=0 \\ \frac{x-2}{3}+\frac{y+4}{3}+\frac{z-3}{3} &=\frac{2}{3} \end{aligned} $$

Step-by-Step Solution

Answer

The system has infinite solutions in the form \( (x, y, z) = (a, b, 3-a-b) \).

Step 1: Simplify the Second Equation

Given the equation \( \frac{x-1}{2} + \frac{y-3}{2} + \frac{z+1}{2} = 0 \), multiply every term by 2 to eliminate the denominators:\[(x-1) + (y-3) + (z+1) = 0 \]Simplifies to:\[x + y + z - 3 = 0 \]Thus, the equation simplifies to \( x + y + z = 3 \), which is the same as the first equation.

Step 2: Simplify the Third Equation

Take the equation \( \frac{x-2}{3} + \frac{y+4}{3} + \frac{z-3}{3} = \frac{2}{3} \). Multiply all terms by 3:\[(x-2) + (y+4) + (z-3) = 2 \]Simplify:\[x + y + z - 1 = 2 \]Which further simplifies to \( x + y + z = 3 \), coinciding again with the first equation.

Step 3: Analyze the System of Equations

All three equations simplify to \( x + y + z = 3 \). This indicates dependence in the system, meaning solution cannot be determined for unique \( x, y, z \) without additional information.

Step 4: Parametrize the Solution

We parametrize the variables assuming one variable can be expressed in terms of others. Let’s choose \( x = a \), \( y = b \). Then,\[z = 3 - a - b\]This provides a parametric solution for all \( a, b \) satisfying the equation.

Key Concepts

Parametric Solution

A parametric solution involves expressing variables in a system of equations in terms of one or more parameters. These parameters are freely chosen variables that allow us to describe a set of solutions instead of a single answer. In this exercise, we see that all three equations simplify into the same form: \( x + y + z = 3 \). This redundancy in the equations implies that any unique solution must be expressed in terms of free parameters.
Let’s choose \( x = a \) and \( y = b \) as our parameters. This gives us the flexibility to calculate \( z \) as \( z = 3 - a - b \).
With this approach:

• \( x \) can be any real number denoted by \( a \).
• \( y \) can be any real number denoted by \( b \).
• \( z \) is then dependent on the choices of \( a \) and \( b \).

This method is useful when the system has fewer independent equations than unknowns, leading to an infinite number of solutions.

Elimination Method

The elimination method is a process used to eliminate variables in a system of equations, transforming it into a simpler equivalent system. The goal is to solve the resulting system with minimal variables. In the given exercise, we utilized multiplication to eliminate fractional coefficients and then simplify the equations further.
In detail:

• Second Equation: Multiplied by 2, it simplified to \( x + y + z = 3 \).
• Third Equation: Multiplied by 3, also simplified to \( x + y + z = 3 \).

By these operations, you'll notice each equation reduced to the same linear equation. This demonstrated that the system is "dependent," because the equations provide the same information about \( x, y, \) and \( z \). Thus, the elimination method confirmed that not all the original equations were needed independently.

Dependent System

A dependent system of equations is one in which the equations are not independent of each other. Instead, they describe the same constraint or provide redundant information. Here, all equations in the system led to \( x + y + z = 3 \), revealing the system's dependency.
For understanding, it's useful to recognize features of dependent systems:

• They have infinite solutions because they represent the same planar condition in the space.
• There is no unique solution without additional constraints.
• Each equation is a scaled version or a linear combination of another.

To identify a dependent system, simplifying or eliminating variables often reveals repeating patterns like identical equations. In practice, recognizing dependency can help efficiently tackle complex systems by discerning unnecessary redundancies.